1. ## Help with groups

Hi
My wife is trying to do the question below. I cannot help out and wondered if someone could take me through it.
Thanks

2. ## Re: Help with groups

Originally Posted by Roger12
Hi
My wife is trying to do the question below. I cannot help out and wondered if someone could take me through it.
Because $a+b+1=b+a+1$ we see $a*b=b*a$ so it is commutative.
$(a*b)*c=(a*b)+c+1=(a+b+1)+c+1=a+(b+c+1)+1=a*(b*c)$ so it is associative.
You wife needs to show that $-1$ is the operational identity.
You wife needs to show that $-a-2$ is the operational inverse of $a$.

3. ## Re: Help with groups

verifying that -1 is the identity is rather easy one you "know" -1 is the identity. but how did Plato know it was?

it involves a bit of "detective work". let's find out what the identity would have to be, if there actually is one.

by definition, an identity for * is an element e of Z, with:

a*e = e*a = a, for any element a of Z.

since we know * is commutative (because of Plato's post above), it suffices to find e with a*e = a, for all a (in Z).

so we write down an equation in Z, and "solve for e".

if a*e = a, what this means in terms of the operations we're used to is:

a+e+1 = a.

we can, once we're "back in the ordinary integers", subtract a from both sides, to get:

e+1 = 0, which makes it clear e = -1 (since that's the only integer e with e+1 = 0, or: if you like, subtract 1 from both sides).

we can use "the same trick" to find out what a-1 has to be.

by definition, a-1 is an element b of Z with a*b = b*a = e.

now, we know what "e" is (see above), so we do pretty much what we did above: write a*b = e in terms of regular operations in Z, and "solve for b".

a*b = e translates into:

a+b+1 = -1

subtract 1 from both sides:

a+b = -2

subtract a from both sides:

b = -2-a (since this formula only depends on "a", we're good).

my point being, it's often not profitable to just "guess" what the identity and/or inverse might be. often, with just a little bit of reasoning, we can figure out what they HAVE to be.